second difference is constant so it could defined by quadratic formula, u(x) = ax^2 + bx + c because 2a = Second difference = 2 so a = 1 then u(x) = x^2 + bx + c
build equations from known values to find b and c for x=1, u(1) = 5 = 1^2 + b(1) + c b + c = 4 ...(1)
x=2, u(2) = 12 = 2^2 + b(2) + c 2b + c = 8 ... (2)
solve b = 4, c =0 so u(x) = x^2 + 4x ... formula for n term (i use x not n here)
difference between consecutive term is 245, we have 245 = u(x+1) - u(x) 245 = (x+1)^2 + 4(x+1) - (x^2 + 4x) 245 = x^2 + 2x + 1 + 4x + 4 - x^2 - 4x 245 = 2x + 5 x = 120 it's between term 120 and 121